Integrating with Inverse Trigonometric Functions
Running the standard derivatives backwards gives two integrals you'll use constantly.
Integrating to give inverse trig functions
Reversing the derivatives above (with a scaling constant $a>0$) gives two integrals worth memorising:
$$\int\frac1{a^2+x^2}\,dx=\frac1a\arctan\left(\frac xa\right)+c$$
$$\int\frac1{\sqrt{a^2-x^2}}\,dx=\arcsin\left(\frac xa\right)+c, \; |x|\lt a$$
Compare $16+x^2$ with $a^2+x^2$: here $a=4$.
$\displaystyle\int\frac9{16+x^2}\,dx=9\int\frac1{16+x^2}\,dx=9\times\frac14\arctan\left(\frac x4\right)+c=\frac94\arctan\left(\frac x4\right)+c$.
If the coefficient of $x^2$ isn't $1$, factor it out before matching to the standard form — e.g. $9+4x^2=4\left(\dfrac94+x^2\right)$, so $a=\dfrac32$ and there's an overall factor of $\dfrac14$ to carry through the integral.
Split the numerator: $\displaystyle\int\frac x{\sqrt{4-x^2}}\,dx+3\int\frac1{\sqrt{4-x^2}}\,dx$.
For the first integral, use the substitution $u=4-x^2$, $du=-2x\,dx$: $\displaystyle\int\frac x{\sqrt{4-x^2}}\,dx=-\frac12\int u^{-1/2}\,du=-\sqrt u+c=-\sqrt{4-x^2}+c$.
For the second, $a=2$: $3\displaystyle\int\frac1{\sqrt{4-x^2}}\,dx=3\arcsin\left(\frac x2\right)+c$.
Combining: $\displaystyle\int\frac{x+3}{\sqrt{4-x^2}}\,dx=-\sqrt{4-x^2}+3\arcsin\left(\frac x2\right)+c$.
Exam-style questions
Question 1 6 marks
Find $\displaystyle\int\frac5{9+25x^2}\,dx$.
Factor: $9+25x^2=25\left(\dfrac9{25}+x^2\right)$, so with $a=\dfrac35$: $\displaystyle\int\frac1{9+25x^2}\,dx=\frac1{25}\int\frac1{\frac9{25}+x^2}\,dx$. (M1, A1)
$=\dfrac1{25}\times\dfrac1{3/5}\arctan\left(\dfrac x{3/5}\right)+c=\dfrac1{15}\arctan\left(\dfrac{5x}3\right)+c$. (M1, A1)
So $\displaystyle\int\frac5{9+25x^2}\,dx=5\times\dfrac1{15}\arctan\left(\dfrac{5x}3\right)+c=\dfrac13\arctan\left(\dfrac{5x}3\right)+c$. (A1, A1)
Question 2 6 marks
Evaluate $\displaystyle\int_0^1\frac1{\sqrt{4-x^2}}\,dx$, leaving your answer in terms of $\pi$.
With $a=2$: $\displaystyle\int\frac1{\sqrt{4-x^2}}\,dx=\arcsin\left(\frac x2\right)+c$. (M1, A1)
$\Big[\arcsin\left(\dfrac x2\right)\Big]_0^1=\arcsin\left(\dfrac12\right)-\arcsin(0)=\dfrac\pi6-0=\dfrac\pi6$. (M1, A1, A1, A1)